Suppose the number of radios in a household has a binomial distribution with parameters n=11 and p=40%. Find the probability of a household having: a. 1 or 9 radios b. 7 or fewer radios c. 5 or more radios d. fewer than 9 radios e. more than 7 radios

Step-by-step explanation:

The formula for binomial distribution is expressed as

P (x=r) = nCr * q^ (n-r) * p^r

From the information given,

n = 11

p = 40% = 40/100 = 0.4

q = 1 - 0.4 = 0.6

x represent the number of radios

a) P (x = 1) or P (x = 9)

P (x = 1) = 11C1 * 0.6^ (11-1) * 0.4^1

P (x = 1) = 0.027

P (x = 9) = 11C9 * 0.6^ (11-9) * 0.4^9

P (x = 9) = 0.0052

P (x = 1) or P (x = 9) = 0.027 + 0.0052 = 0.0322

b) P (x lesser than or equal to 7) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6)

P (x = 0) = 11C0 * 0.6^ (11-0) * 0.4^0 = 0.004

P (x = 1) = 0.027

P (x = 2) = 11C2 * 0.6^ (11-2) * 0.4^2 = 0.089

P (x = 3) = 11C3 * 0.6^ (11-3) * 0.4^3 = 0.177

P (x = 4) = 11C4 * 0.6^ (11-4) * 0.4^4 = 0.24

P (x = 5) = 11C5 * 0.6^ (11-5) * 0.4^5 = 0.22

P (x = 6) = 11C6 * 0.6^ (11-6) * 0.4^6 = 0.15

P (x = 7) = 11C7 * 0.6^ (11-7) * 0.4^7 = 0.15

P (x lesser than or equal to 7) = 0.004 + 0.027 + 0.089 + 0.177 + 0.24 + 0.22 + 0.15 + 0.07 = 0.977

c) P (x greater than or equal to 5) = 1 - P (x lesser than or equal to 4) = 1 - (0.004 + 0.027 + 0.089 + 0.177 + 0.24) = 1 - 0.537 = 0.463

d) P (x lesser than 9) = P (x lesser than or equal to 7) + P (x = 8)

P (x = 8) = 11C8 * 0.6^ (11-8) * 0.4^8 = 0.02

P (x lesser than 9) = 0.977 + 0.02 = 0.997

e. P (x greater than 7) = 1 - P (x lesser than or equal to 7) = 1 - 0.977 = 0.023